I would like some guidance of how to create a power supply for a 6 volt pump ,which has a resistance of 9 ohms, 1 watt . I was able to build a voltage divider circuit R1 5K, R2 10 K , gave me 5.74vdc with a supply of 9vdc. If I plug the 6 volt pump directly into the wallart it will work .The wallart output is 9vdc 200ma, input is 120Vac 4.4 watts . So I can see the problem , the supply is probably too small to supply enough power to do this .I could just keep trying other power supplies but that don't make it right.. Is there a formula someone could share or shed some light on what I need to do to do this properly. Thank you

You can’t use a resistor divider to create 6V from 9V and then plug you pump into the middle of the divider; there’s very little current available at the middle of the resistor divider to actually power the pump.

Do you have specs on the pump (e.g. datasheet)? Do you need to vary its speed or just switch it on and off?

So I found a formula

I=PV

P=1,V=9

IR1=1/9=.11111111Amps

R1(VR!)

VB=9v,VL=6v

VR!=9-6

VR!=3v

So R1=3/.11111111A

R1=27 Ohms

PR!=VR!xIR1

3x.11111111=.3 watts

Looks like 27ohm resistor at 1/2 watt should work fine for this. Any thoughts here as to viability?

Amp meter shows about.08 amps .

A voltage regulator (or the correct power supply) is the correct solution.

Your problem will be that **the motor/pump resistance varies with the load.** 9 Ohms is probably the minimum (maximum current & maximum wattage) when it's stalled or just-starting. With the motor running you'll end up with more voltage than calculated applied to the motor. If you in

crease the series resistance to correct for that, the motor might not start (because the voltage is too low before the motor starts running.

6V across 9 Ohms is 0.67 (670mA) Amps and 4 Watts. **You're also exceeding the 200mA rating on the wall wart.**

You want 3V dropped across a series resistor and with 0.76 Amps that's 4.5 Ohms. (Or half of voltage that's across the motor so half the motor's resistance.)

**P.S.**

I'm violating the *"never do math in public"* rule, so please double-check my numbers. And, then go buy the correct power supply!

Darn, I thought I figured it out for myself, and yes I know the right power supply would be the better option, was just trying to use what I had.

The resistor in series is a terrible idea, since it will drop the voltage as more current is drawn. You need to build a real power supply, say using a 7806 or some other regulated power supply chip. That will provide 6 volts, whether at low or high (within the limits of the chip) current draw. 1 watt at 6 v. should be 0.167 amps, well within the 78xx family current limit. But a heat sink would still be a good idea. Study up on them. They're very simple to apply.

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